hw5.sol.Spring07

# hw5.sol.Spring07 - ME 452: Machine Design II Solution to...

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ME 452: Machine Design II Solution to Homework Set 5. Problem 1. (Problem 7-1, page 388). A shaft is to be designed to an infinite life fatigue factor of safety, nf = 2.0, with the following specifications: Ma = 600 in lb Ta = 400 in lb Mm = 500 in lb Tm = 300 in lb Kf = 2.2 Kfs = 1.8 The material properties are specified as Sut = 100,000 psi, Sy = 80,000 psi, and Se = 30,000 psi. Find the appropriate shaft diameter using (a) DE-Gerber, (b) DE-Elliptic, (c) DE-Soderberg, and (d) DE-Goodman. Compare and discuss results. Solution: The loads, stress concentration factors, and material properties are all give, so the solution process starts with the equations in Chapter 7. Equations 7-8 through 7-14 represent the solution of the material of chapter 6 on fatigue, using the four failure theories above, and solving for the shaft diameter, d. Though we could use the equations from Chapter 6 and solve them ourselves for d, this has been done for us in Chapter 7, and we will take advantage of this. (a) For the DE-Gerber solution, use Equation (7-70) page 357: 1/3 1/2 2 8 2 11 f e eu t nA BS d SA S π ⎛⎞ ⎡⎤ ⎜⎟ ⎢⎥ =+ + ⎝⎠ ⎣⎦ ⎩⎭ where () ( ) ( ) 22 2 2 43 fa f s a fm f s m AK M K T BK M K T Substituting in the given values gives: 4 2.2 600 3 1.8 400 4 2.2 500 3 1.8 300 AB + × + × Therefore A = 2920 in lb B = 2391 in lb Then 2 8 2 2920 2 2391 30000 1 1 1.016 30,000 2920 100,000 di n ⎧⎫ ×× × × ⎪⎪ + = ⎨⎬

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(b) For the DE-Elliptic solution, use Equation (7-12) page 357. 1/3 1/2 22 16 43 ff a f s a f m f s m ee y y nK M K T K M K T d SSSS π ⎧⎫ ⎡⎤ ⎛⎞ ⎪⎪ ⎢⎥ =+ + + ⎜⎟ ⎨⎬ ⎝⎠ ⎣⎦ ⎩⎭ Substituting in the given values gives: 16 2 2.2 600 1.8 400 2.2 500 1.8 300 4343 30,000 30,000 80,000 80,000 d ×× × × × + + Therefore d = 1.012 in (c) For the DE-Soderberg solution, use Equation (7-14) page 357: 16 a f s a f m f s m M K T K M K T d SS + + Substituting in the given values: 16 2 2.2 600 1.8 400 2.2 500 1.8 300 30,000 30,000 80,000 80,000 d × × × + + Therefore d = 1.09 in (d) For the DE-Goodman solution, use Equation (7-8) page 356, which is exactly the same as Equation (7-14) but with S ut substituted in for S y : 2 2 16 a f s a f m f s m e e ut ut M K T K M K T d + + Substituting in the given values: 16 2 2.2 600 1.8 400 2.2 500 1.8 300 30,000 30,000 100,000 100,000 d × × × + + Therefore,
d = 1.073 in Discussion: The four failure theories are plotted below on an alternating-stress, midrange-stress axis system. Note that Soderberg and Goodman are straight lines, Gerber is a parabola, and the Elliptic equation is an ellipse. Note also how close the failure lines are for Gerber and the Elliptic equation. Also plotted in the figure is a load line. This load line is obtained by calculating the midrange and alternating stresses for a 1.0 inch diameter shaft with the loads as given above. These stresses are calculated using Equations (7-5) and (7-6) page 356: 1/2 22 33 32 16 3 fm f s m m KM KT dd σ ππ ⎡⎤ ⎛⎞ ⎢⎥ =+ ⎜⎟ ⎝⎠ ⎣⎦ 32 16 3 fa f

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## This note was uploaded on 02/09/2012 for the course ME 352 taught by Professor Staff during the Fall '08 term at Purdue.

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hw5.sol.Spring07 - ME 452: Machine Design II Solution to...

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