6_ch 09 Mechanical Design budynas_SM_ch09

# 6_ch 09 Mechanical Design budynas_SM_ch09 - = Mr y J =...

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244 Solutions Manual Instructor’s Solution Manual to Accompany Mechanical Engineering Design Weldment Specifications: Pattern: All-around square Electrode: E6010 Type: Two parallel fillets Ans. Two transverse fillets Length of bead: 12 in Leg: 1 / 4 in For a figure of merit of, in terms of weldbead volume, is this design optimal? 9-12 Decision: Choose a parallel fillet weldment pattern. By so-doing, we’ve chosen an optimal pattern (see Prob. 9-9) and have thus reduced a synthesis problem to an analysis problem: Table 9-1: A = 1 . 414 hd = 1 . 414 ( h )( 3 ) = 4 . 24 h in 3 Primary shear τ y = V A = 3000 4 . 24 h = 707 h Secondary shear Table 9-1: J u = d (3 b 2 + d 2 ) 6 = 3[3(3 2 ) + 3 2 ] 6 = 18 in 3 J = 0 . 707( h )(18) = 12 .
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Unformatted text preview: = Mr y J = 3000(7 . 5)(1 . 5) 12 . 7 h = 2657 h = τ ±± y τ max = ± τ ±± 2 x + ( τ ± y + τ ±± y ) 2 = 1 h ² 2657 2 + (707 + 2657) 2 = 4287 h Attachment (1018 HR): S y = 32 kpsi, S ut = 58 kpsi Member (A36): S y = 36 kpsi The attachment is weaker Decision: Use E60XX electrode τ all = min[0 . 3(58), 0 . 4(32)] = 12 . 8 kpsi τ max = τ all = 4287 h = 12 800 psi h = 4287 12 800 = . 335 in Decision: Specify 3 / 8 " leg size Weldment Speciﬁcations: Pattern: Parallel ﬁllet welds Electrode: E6010 Type: Fillet Ans. Length of bead: 6 in Leg size: 3 / 8 in...
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