hw5.sol.Fall.2010x

hw5.sol.Fall.2010x - 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 400). A shaft is to be designed to an infinite life fatigue factor of safety, n f = 2.0, with the following specifications: M a = 70 N-m T a = 45 N-m M m = 55 N-m T m = 35 N-m K f = 2.2 K fs = 1.8 The material properties are specified as S ut = 700 MPa, S y = 560 MPa, and S e = 210 MPa. 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. To help with the units, note that keeping all moments in terms of N and mm will make stresses come out in MPa. So first let’s convert: M a = 70,000 N-mm T a = 45,000 N-mm M m = 55,000 N-mm T m = 35,000 N-mm (a) For the DE-Gerber solution, use Equation (7-10) page 369: 1/3 1/ 2 2 8 2 1 1 f e e ut n A BS d S AS π = + + where ( 29 ( 29 ( 29 ( 29 2 2 2 2 4 3 4 3 f a fs a f m fs m A K M K T B K M K T = + = + Substituting in the given values gives: ( 29 ( 29 ( 29 ( 29 2 2 2 2 4 2.2 70,000 3 1.8 45,000 4 2.2 55,000 3 1.8 35,000 A B = × + × = × + ×
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Therefore A = 338,400 N-mm B = 265,500 N-mm Then 1/3 1/2 2 8 2 338,400 2 265,500 210 1 1 25.85 mm 210 338,400 700 d π × × × × = + + = × × (b) For the DE-Elliptic solution, use Equation (7-12) page 369. 1/3 1/ 2 2 2 2 2 16 4 3 4 3 f f a fs a f m fs m e e y y n K M K T K M K T d S S S S = + + + Substituting in the given values gives: 1/3 1/2 2 2 2 2 16 2 2.2 70,000 1.8 45,000 2.2 55,000 1.8 35,000 4 3 4 3 210 210 560 560 d × × × × × = + + + Therefore d = 25.77 mm (c) For the DE-Soderberg solution, use Equation (7-14) page 369: 1/3 1/ 2 1/ 2 2 2 2 2 16 4 3 4 3 f f a fs a f m fs m e e y y n K M K T K M K T d S S S S = + + + Substituting in the given values:
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hw5.sol.Fall.2010x - ME 452: Machine Design II Solution to...

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