8_ch 08 Mechanical Design budynas_SM_ch08

8_ch 08 Mechanical Design budynas_SM_ch08 -...

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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 ...
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