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3_ch 09 Mechanical Design budynas_SM_ch09

# 3_ch 09 Mechanical Design budynas_SM_ch09 -...

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Chapter 9 241 Maximum shear τ max = τ 2 x + ( τ y + τ y ) 2 = F 2 . 97 2 + (0 . 556 + 2 . 97) 2 = 4 . 61 F kpsi F = τ all 4 . 61 Ans. which is twice τ max / 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 mm 2 Members’endurance limit: AISI 1010 steel S ut = 320 MPa, S e = 0 . 5(320) = 160 MPa k a = 272(320) 0 . 995 = 0 . 875 k b = 1 (direct shear) k c = 0 . 59 (shear) k d = 1 k f = 1 K f s = 1 2 . 7 = 0 . 370 S se = 0 . 875(1)(0 . 59)(0 . 37)(160) = 30 . 56 MPa Electrode’s endurance: 6010 S ut = 62(6 . 89) = 427 MPa S e = 0 . 5(427) = 213 . 5 MPa k a = 272(427)
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