3_ch 09 Mechanical Design budynas_SM_ch09

3_ch 09 Mechanical Design budynas_SM_ch09 -...

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Unformatted text preview: budynas_SM_ch09.qxd 12/01/2006 16:16 FIRST PAGES Page 241 Chapter 9 Maximum shear τmax = F= τx 2 + ( τ y + τ y ) 2 = F 2.972 + (0.556 + 2.97) 2 = 4.61 F kpsi τall 4.61 Ans. which is twice τmax /9.22 of Prob. 9-5. 9-7 Weldment, subjected to alternating fatigue, has throat area of A = 0.707(6)(60 + 50 + 60) = 721 mm2 Members’ endurance limit: AISI 1010 steel Sut = 320 MPa, Se = 0.5(320) = 160 MPa ka = 272(320) −0.995 = 0.875 kb = 1 (direct shear) kc = 0.59 (shear) kd = 1 kf = 1 1 = = 0.370 Kfs 2.7 Sse = 0.875(1)(0.59)(0.37)(160) = 30.56 MPa Electrode’s endurance: 6010 Sut = 62(6.89) = 427 MPa Se = 0.5(427) = 213.5 MPa ka = 272(427) −0.995 = 0.657 kb = 1 (direct shear) kc = 0.59 (shear) kd = 1 k f = 1/ K f s = 1/2.7 = 0.370 . Sse = 0.657(1)(0.59)(0.37)(213.5) = 30.62 MPa = 30.56 Thus, the members and the electrode are of equal strength. For a factor of safety of 1, Fa = τa A = 30.6(721)(10−3 ) = 22.1 kN Ans. 241 ...
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This note was uploaded on 12/13/2011 for the course EML 3013 taught by Professor Shingley during the Fall '11 term at UNF.

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