8_ch 09 Mechanical Design budynas_SM_ch09

# 8_ch 09 Mechanical Design budynas_SM_ch09 - Member(A36 S y...

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246 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design τ max = ± ( τ ±± x + τ ± x ) 2 + τ ±± 2 y = 1 h ² (2651 + 707 . 5) 2 + 2651 2 = 4279 h psi Relate stress and strength: τ max = τ all 4279 h = 12 800 h = 4279 12 800 = 0 . 334 in 3 / 8in Weldment Speciﬁcations: Pattern: Horizontal parallel weld tracks Electrode: E6010 Type of weld: Two parallel ﬁllet welds Length of bead: 6 in Leg size: 3 / 8 in Additional thoughts: Since the round-up in leg size was substantial, why not investigate a backward C ± weld pattern. One might then expect shorter horizontal weld beads which will have the advan- tage of allowing a shorter member (assuming the member has not yet been designed). This will show the inter-relationship between attachment design and supporting members. 9-14 Materials:
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Unformatted text preview: Member (A36): S y = 36 kpsi, S ut = 58 to 80 kpsi; use S ut = 58 kpsi Attachment (1018 HR): S y = 32 kpsi, S ut = 58 kpsi τ all = min[0 . 3(58), 0 . 4(32)] = 12 . 8 kpsi Decision: Use E6010 electrode. From Table 9-3: S y = 50 kpsi, S ut = 62 kpsi, τ all = min[0.3(62), 0 . 4(50)] = 20 kpsi Decision: Since A36 and 1018 HR are weld metals to an unknown extent, use τ all = 12 . 8 kpsi Decision: Use the most efﬁcient weld pattern–square, weld-all-around. Choose 6 " × 6 " size. Attachment length: l 1 = 6 + a = 6 + 6 . 25 = 12 . 25 in Throat area and other properties: A = 1 . 414 h ( b + d ) = 1 . 414( h )(6 + 6) = 17 . h ¯ x = b 2 = 6 2 = 3 in, ¯ y = d 2 = 6 2 = 3 in...
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