Problem 6-62
description
Transcript of Problem 6-62
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Problem 6-62A carton turnover machine is shown below.
16 in
15 in
7.3 in
7 in
600
2 in3 in
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Problem 6-62At the instant shown, the 15 in link is driven clockwise at 5 rad/sec, and decelerating at
40 rad/sec2. Determine the instantaneous torque required to operate the 15 in arm, and the forces at the two lower bearings. The two long links closely resemble slender members, made from steel, with a width of 1 in and a thickness of
0.25 in. The carrier and carton weigh 26 lb and its mass moment of inertia, relative to an axis through its center, is 2.75 lb in s2.
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Kinematic Diagram
2
A
B
C
D
G2
G4
G3 3
4
310
7.3”
16”
15”5.83”
600
3.6
220
7
56.30
2
3
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Velocity Analysis
VB = 75 in/s
VB = 2 rAB = (5 rad/s)(15 in) = 75 in/s
VC/B
VC
VC = VB +> VC/B
to CB
to CDVG3/BVG3/C
VG3
to G3C
to G3B
VG3 = VB +> VG3/B
= VC +> VG3/C
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Velocity Results
• VC = 40.4 in/s 680
• VC/B = 49.8 in/s
• VX/B = 25.6 in/s 56.30
• VX/C = 41.5 in/s 310
• VG3 = 77.7 in/s 490
• VG2 = 37.5 in/s 300
• VG4 = 20.2 in/s 680
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Acceleration Analysis
• anC = (40.4)2/16 = 102.0 in/s2 22.00
• anG3/C = (41.5)2/5.83 = 295.4 in/s2 59.00
• anG3/B = (25.6)2/3.6 = 182.0 in/s2 33.70
• atB = (15)(40) = 600.0 in/s2 30.00
• anC/B = (49.8)2/7 = 354.3 in/s2
• anB = (75)2/15 = 375.0 in/s2 60.00
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Acceleration Analysis
anC+>at
C= anB+>at
B+> anC/B+>at
C/B
to CB to CD
anC
anB
atB
anC/B
atC
atC/B
aB
aC
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Acceleration Analysis• at
C = 250.8 in/s2 68.00
• =(250.8)/16 =15.68 rad/s2 ccw
• =(332.7)/7 =47.53 rad/s2
ccw
• aC = 270.7 in/s2
• atC/B = 332.7 in/s2
• aG4 = 135.4 in/s2
• aB = 707.5 in/s2 620
• aG2 = 353.8 in/s2
620
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anC
anB
atB
anC/B
atC
atC/B
anG3/B an
G3/C
atG3/B at
G3/C
to G3C
to G3BaG3
• aG3 = 672 in/s2 830
aG3= aB+>anG3/B+>at
G3/B
= aC+> anG3/C+>at
G3/C
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Inertia Forces: Link 2
15 in600
• aG2 = 353.8 in/s2
620
• Vol2 = (16 in)(1 in)(.25 in) = 4 in3
• W2 = (.283 lb/in3)(4 in3) = 1.13 lb
• FiG2 = (1.13 lb)(353.8 in/s2)/(386.4 in/s2)
= 1.03 lbs 620
FiG2
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Inertia Forces: Link 2
15 in600
FiG2
• IG2 = ¼ [(1.13 lb)/(386.4 in/s2)](16 in)2
= 0.19 lb in s2
• = 40 rad/s2 ccw
• TiG2 = (0.19 lb in s2)(40 rad/s2)
= 7.5 lb in cw
TiG2
• Assume slender rod
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Inertia Forces: Link 3
• aG3 = 672 in/s2 830
• =47.53 rad/s2ccw
• FiG3 = (26 lb)(672 in/s2)/(386.4 in/s2)
= 45.2 lbs 830
FiG3
• TiG3 = (2.75 in lb s2)(47.53 rad/s2)
= 130.7 in lbs cw
TiG3
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• Vol4 = (17 in)(1 in)(.25 in) = 4.25 in3
• W4 = (.283 lb/in3)(4.25 in3) = 1.20 lb
Inertia Forces: Link 4
16 in220
• aG4 = 135.4 in/s2
• FiG4 = (1.20 lb)(135.4 in/s2)/(386.4 in/s2)
= 0.42 lb
FiG4
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16 in220
Inertia Forces: Link 4
FiG4
• IG4 = ¼ [(1.20 lb)/(386.4 in/s2)](17 in)2
= 0.22 lb in s2
• TiG4 = (0.22 lb in s2)(15.68 rad/s2)
= 3.4 lb in cw
TiG4
• Assume slender rod
• = 15.68 rad/s2
ccw
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Free Body Diagrams
Cx
Cy
Cy
Cx
Dy
Dx
Ay
Ax
Bx
ByBy
Bx
FiG3
TiG3
FiG2
TiG4
FiG2
TiG2
W3
W4
W2
TA
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FBD: Link 2
7.5 600
Ay
Ax
By
Bx
FiG2
TiG2
620
15 in
Fx = 0
Ax - Bx + FiG2 cos62 = 0
Fy = 0
Ay - By +FiG2 sin62 - W2 = 0
W2
A = 0
Bx[15(sin60)] + By[15(cos60)] + W2[7.5(cos60)] - TiG2 - TA
- [FiG2(sin62)][7.5(cos60)] – [Fi
G2(cos62)][7.5(sin60)] = 0
TA
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FBD: Link 3
Bx
Cx
FiG3
3 in
7 in
2 in
Cy
By
TiG3
W3
830
Fx = 0
Bx + Cx + FiG3 cos83 = 0
Fy = 0
By + Cy + FiG3 sin83 – W3 = 0
c = 0
W3(3) - Bx(7) - (FiG3 sin83)(3) - (Fi
G3 cos83)(5) - TiG3 = 0
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FBD: Link 4
16 in
220
Dy
Dx
Cy
Cx FiG4
TiG4
W4
8
Fx = 0
Dx - Cx = 0
Fy = 0
Dy - Cy + FiG4 - W4 = 0
D = 0
Cx[16(sin22)] + Cy[16(cos22)] + W4[8(cos22)]
- FiG4[8(cos22)] - Ti
G4 = 0
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Finally, Solving the Nine Simultaneous Equations:
• Ax = -23.3 lbs• Ay = -11.5 lbs• Bx = -22.8 lbs• By = -11.7 lbs
• Cx = 17.3 lbs• Cy = -7.2 lbs• Dx = 17.3 lbs• Dy = -6.4 lbs
• TA = -394.1 in lbs