hw6csol.sp12 - Name of Student _ ME 452: Machine Design II...

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1 Name of Student ______________________ ME 452: Machine Design II Spring Semester 2012 Homework Set 6 Attempt by Wednesday, February 22nd Solve the following problems from Chapter 7, Shigley’s Mechanical Engineering Design, Ninth Edition, R.G. Budynas and J.K. Nisbett. 1. Problem 7-1 page 400. 2. Problem 7-4 page 401. 3. Problem 7-17 page 402.
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2 Solution to Homework Set 6. Problem 7-1. (a) For the DE-Gerber, see Eq. (7-10), page 369, the coefficients are () ( ) 22 4 3 4 2.2 70 3 1.8 45 338.4 fa f s a AK M K T =+ = × + × = N.m (1a) and ( ) 4 3 4 2.2 55 3 1.8 35 265.5 ff s mm BK M K T = × + × = N.m (1b) Substituting Equations (1a) and (1b) into Equation (7-10), page 369, the diameter of the shaft is 1/3 1/2 2 6 66 8 2 338.4 2 265.5 210 10 11 210 10 338.4 700 10 d π ⎧⎫ ⎛⎞ ⎡⎤ ×× × × × ⎪⎪ ⎜⎟ + ⎨⎬ ⎢⎥ ⎣⎦ ⎝⎠ ⎩⎭ (2a) that is 3 25.85 0 m 1 d or 25. mm 85 d = (2b) The DE-elliptic, see Eq. (7-12), page 369, can be written as 1/3 16 ey nA B d SS (3) Substituting Equations (1a) and (1b) into Equation (3), the diameter of the shaft is 1/3 16 2 338.4 265.5 210 10 560 10 d × (4a) that is 3 25.77 0 m 1 d or 25. mm 77 d = (4b) The DE-Soderberg, see Eq. (7-14), page 369, can be written as 1/3 16 d (5) Substituting Equations (1a) and (1b) into Equation (5), the diameter of the shaft is 1/3 16 2 338.4 265.5 210 10 560 10 d × (6a) that is 3 27.70 0 m 1 d or 27. mm 70 d = (6b)
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3 The DE-Goodman, see Eq. (7-8), page 368, can be written as 1/3 16 ut e nA B d SS π ⎛⎞ =+ ⎜⎟ ⎝⎠ (7) Substituting Equations (1a) and (1b) into Equation (7), the diameter of the shaft is 1/3 66 16 2 338.4 265.5 210 10 700 10 d × ×× (8a) that is 3 27.27 0 m 1 d or 27. mm 27 d = (8b) A summary and a comparison of the four answers for the diameter of the shaft are presented in the following table. Criterion d (mm) Compared to DE Gerber DE Gerber 25.85 DE Elliptic 25.77 0.31% Lower Less conservative DE Soderberg 27.7 7.2% Higher More conservative DE Goodman 27.27 5.5% Higher More conservative Problem 7-4. We have a design task of identifying bending moment diagrams and torsion diagrams which are preliminary to an industrial roller shaft design. Let point C represent the center of the span of the roller. The y and Z components of the force at point C, respectively, are 30 8 240 y C F = lb and 0.4 240 96 z C F = ×= lb (1) The torque is 2 96 2 192 z C TF = × = lb.in (2)
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4 Therefore, the z-component of the force at point B is 192 128 1.5 1.5 z B T F == = lb (3) The y-component of the force at point B is tan 20 128tan 20 46.6 yz BB FF = lb (4) (a) The free body diagram in the xy-plane is shown in the figure below. The sum of the moments about point O can be written as 240 5.75 11.5 46.6 14.25 0 A O y MF ×− × = (5) The y-component of the force at point A is 240 5.75 46.6 14.25 62.3 11.5 y A F ×−× lb (6) The sum of the moments about point A can be written as 11.5 46.6 2.75 240 5.75 0 y AO × × = (7)
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hw6csol.sp12 - Name of Student _ ME 452: Machine Design II...

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