8_ch 08 Mechanical Design budynas_SM_ch08

# 8_ch 08 Mechanical Design budynas_SM_ch08 -...

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

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: budynas_SM_ch08.qxd 11/30/2006 15:49 Page 211 FIRST PAGES 211 Chapter 8 Eq. (8-23), from the Wileman et al. ﬁnite element study, using the general expression, km Ed 8-19 ( d / l ) =0.5 = 0.789 52 exp[0.629 14(0.5)] = 1.08 For cast iron, from Table 8-8: A = 0.778 71, B = 0.616 16, E = 14.5 Mpsi km = 14.5(106 )(0.625)(0.778 71) exp 0.616 16 0.625 1.5 = 9.12(106 ) lbf/in This member’s spring rate applies to both members. We need km for the upper member which represents half of the joint. kci = 2km = 2[9.12(106 )] = 18.24(106 ) lbf/in For steel from Table 8-8: A = 0.787 15, B = 0.628 73, E = 30 Mpsi km = 30(106 )(0.625)(0.787 15) exp 0.628 73 0.625 1.5 = 19.18(106 ) lbf/in ksteel = 2km = 2(19.18)(106 ) = 38.36(106 ) lbf/in For springs in series 1 1 1 1 1 = + = + km kci ksteel 18.24(106 ) 38.36(106 ) km = 12.4(106 ) lbf/in Ans. 8-20 The external tensile load per bolt is P= 1π ( 150) 2 (6)(10−3 ) = 10.6 kN 10 4 Also, l = 40 mm and from Table A-31, for d = 12 mm, H = 10.8 mm. No washer is speciﬁed. L T = 2 D + 6 = 2(12) + 6 = 30 mm l + H = 40 + 10.8 = 50.8 mm L = 60 mm ld = 60 − 30 = 30 mm lt = 45 − 30 = 15 mm Table A-17: Ad = π (12) 2 = 113 mm2 4 At = 84.3 mm2 Table 8-1: Eq. (8-17): kb = 113(84.3)(207) = 466.8 MN/m 113(15) + 84.3(30) Steel: Using Eq. (8-23) for A = 0.787 15, B = 0.628 73 and E = 207 GPa ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online