3_ch 06 Mechanical Design budynas_SM_ch06

3_ch 06 Mechanical Design budynas_SM_ch06 -...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

Ask a homework question - tutors are online