3_ch 06 Mechanical Design budynas_SM_ch06

3_ch 06 Mechanical Design budynas_SM_ch06 -...

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Unformatted text preview: budynas_SM_ch06.qxd 11/29/2006 17:40 FIRST PAGES Page 149 149 Chapter 6 d 7.62 −0.107 = 32 7.62 −0.107 Eq. (6-20): kb = = 0.858 Eq. (6-18): Se = ka kb Se = 0.792(0.858)(355) = 241 MPa Ans. 6-7 For AISI 4340 as forged steel, Se = 100 kpsi Eq. (6-8): a = 39.9, Table 6-2: b = −0.995 ka = 39.9(260) −0.995 = 0.158 Eq. (6-19): 0.75 0.30 Each of the other Marin factors is unity. kb = Eq. (6-20): −0.107 = 0.907 Se = 0.158(0.907)(100) = 14.3 kpsi For AISI 1040: Se = 0.5(113) = 56.5 kpsi ka = 39.9(113) −0.995 = 0.362 kb = 0.907 (same as 4340) Each of the other Marin factors is unity. Se = 0.362(0.907)(56.5) = 18.6 kpsi Not only is AISI 1040 steel a contender, it has a superior endurance strength. Can you see why? 6-8 (a) For an AISI 1018 CD-machined steel, the strengths are 2.5 mm 20 mm 25 mm Eq. (2-17): Sut = 440 MPa ⇒ HB = S y = 370 MPa Ssu = 0.67(440) = 295 MPa Fig. A-15-15: Fig. 6-21: Eq. (6-32): 2.5 D 25 r = = 0.125, = = 1.25, d 20 d 20 qs = 0.94 K f s = 1 + 0.94(1.4 − 1) = 1.376 K ts = 1.4 For a purely reversing torque of 200 N · m K f s 16T 1.376(16)(200 × 103 N · mm) = τmax = π d3 π (20 mm) 3 τmax = 175.2 MPa = τa Se = 0.5(440) = 220 MPa 440 = 129 3.41 ...
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