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Unformatted text preview: 1 Name of Student ______________________ ME 452: Machine Design II Spring Semester 2012 Homework Set 5 Attempt by Wednesday, February 15th Solve the following problems from Chapters 6 and 7, Shigley’s Mechanical Engineering Design, Ninth Edition, R.G. Budynas and J.K. Nisbett. 1. Problem 625 page 350. 2. Problem 656 page 353. 3. Problem 73 page 401. Notes and Reminders Submit the progress report for Project 1 before noon, Room ME 3003, Friday, February 10th. Class Test 1 is in lecture on Friday, February 10th. The test is open book and closed notes and will focus on the material taken from the class lectures and Homework Sets 1, 2, 3, and 4. 2 Solution to Homework Set 5. Problem 625. From Table A20 for AISI 1040 CD steel, see page 1040, the ultimate tensile strength and the yield strength of the steel bar, respectively, are 590 ut S = MPa and 490 y S = MPa (1) Given the maximum and the minimum axial forces 28 max F = kN and 28 min F = − kN then the maximum and the minimum normal stresses, respectively, are 28000 147.4 10 (25 6) max max F A σ = = = × − N/mm 2 147.4 = MPa (2a) and min min 28000 147.4 10 (25 6) F A σ − = = = − × − N/mm 2 147.4 = MPa (2b) The alternating normal stress (excluding the fatigue stress concentration factor) can be written as 2 max min a F F A σ − = ( 3 a ) Then substituting 28 max F = kN and 28 min F = − kN, and the crosssectional geometry of the steel bar, into Equation (4a), the alternating normal stress is 28000 ( 28000) 147.36 MPa 2 10 (25 6) a σ − − = = × × − (3b) Note that the mean normal stress is 2 max min m F F A σ + = = (4) The static factor of safety based on the distortion energy theory (see Chapter 5) can be written as y y a m S n σ σ + = (5a) Substituting Equations (1), (3b), and (4) into Equation (5a), the static factor of safety is 490 3.33 147.36 0 y n + = = (5b) Check. The static factor of safety based upon the distortion energy theory can be written as y y max S n σ = (6a) Substituting Equations (1) and (3) into Equation (6a), the static factor of safety is 490 3.33 147.36 y n = = (6b) 3 To determine the fatigue factor of safety based on infinite life. Using Eq. (68), see page 282, the uncorrected endurance strength of the steel bar is 0.5 0.5 590 295 e u t S S ′ = = × = MPa (7) From Table 62, see page 288, the constants are and 4.51MPa 0.265 a b = = − (8) From Eq. (619), see page 287, the surface factor is 0.265 4.51 590 0.832 b a u t k a S − = = × = (9) From Eq. (621), see page 288, the size factor for axial loading is 1 b k = (10) From Eq. (626), see page 290, the loading factor is 0.85 c k = (11) Using Eq. (618), see page 287, the corrected endurance strength of the solid steel round bar is 0.832 1 0.85 295 208.6 e a b c e S k k k S = ′ = × × × = MPa (12) Using Fig. 620, the notch sensitivity is 0.83 q = (13) Using Figure A151, see page 1026, with the geometrical ratio / . 2 4 d w = (14) the theoretical stress concentration factor is 2.1 t...
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This document was uploaded on 02/22/2012.
 Spring '09
 Machine Design

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